题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1839
思路：先给容量从大到小排序，然后二分，每次都一次spfa一下，判断当前的cost[n]的值。。。
View Code
1 #include<iostream> 2 #include<queue> 3 #include<vector> 4 typedef long long ll; 5 const int MAXN=10000+10; 6 const ll inf=1e60; 7 using namespace std; 8 struct Node{ 9 int v,w,t; 10 }; 11 vector<Node>mp[MAXN]; 12 ll cost[MAXN]; 13 ll weight[MAXN*5]; 14 int n,m,time; 15 int limit; 16 17 18 int cmp(const ll a,const ll b){ 19 return a>b; 20 } 21 22 23 void SPFA(int u){ 24 for(int i=1;i<=n;i++)cost[i]=inf; 25 cost[1]=0; 26 queue<int>Q; 27 Q.push(u); 28 while(!Q.empty()){ 29 int u=Q.front(); 30 Q.pop(); 31 for(int i=0;i<mp[u].size();i++){ 32 int v=mp[u][i].v; 33 int t=mp[u][i].t; 34 int w=mp[u][i].w; 35 //这儿实在是太妙了，每次选的边都有一个限制就行了 36 if(cost[u]+t<cost[v]&&w>=limit){ 37 cost[v]=cost[u]+t; 38 Q.push(v); 39 } 40 } 41 } 42 } 43 44 int main(){ 45 int _case; 46 scanf("%d",&_case); 47 while(_case--){ 48 scanf("%d%d%d",&n,&m,&time); 49 for(int i=1;i<=n;i++)mp[i].clear(); 50 memset(weight,0,sizeof(weight)); 51 for(int i=1;i<=m;i++){ 52 int u,v,w,t; 53 scanf("%d%d%d%d",&u,&v,&w,&t); 54 Node p1,p2; 55 p1.v=u,p2.v=v; 56 p1.t=p2.t=t; 57 p1.w=p2.w=w; 58 weight[i]=w;//用来记录每条路的容量; 59 mp[u].push_back(p2); 60 mp[v].push_back(p1); 61 } 62 sort(weight+1,weight+m+1,cmp); 63 //二分 64 int low=1,high=m; 65 while(low<=high){ 66 int mid=(low+high)/2; 67 limit=weight[mid];//每次都选择一个限制 68 SPFA(1); 69 if(cost[n]==inf||cost[n]>time){ 70 low=mid+1; 71 }else 72 high=mid-1; 73 } 74 printf("%d\n",weight[low]); 75 } 76 return 0; 77 }